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98 ⏐⏐⏐ OHM’S LAW, POWER, AND ENERGY
German (Erlangen,
Cologne)
(1789–1854)
Physicist and
Mathematician
Professor of Physics,
University of
Cologne
+
E
–
FIG. 4.2
Basic circuit.
Courtesy of the
Smithsonian Institution
Photo No. 51,145
In 1827, developed one of the most important laws
of electric circuits: Ohm’s law. When the law was
first introduced, the supporting documentation was
considered lacking and foolish, causing him to lose
his teaching position and search for a living doing
odd jobs and some tutoring. It took some 22 years
for his work to be recognized as a major contribution
to the field. He was then awarded a chair at the
University of Munich and received the Copley
Medal of the Royal Society in 1841. His research
also extended into the areas of molecular physics,
acoustics, and telegraphic communication.
FIG. 4.1
Georg Simon Ohm.
I
+
R V
–
of water flow through the hose, just as applying a higher voltage to the
same circuit will result in a higher current.
Substituting the terms introduced above into Eq. (4.1) results in
Current �
and I � � (amperes, A) (4.2)
E
R �
Equation (4.2) is known as Ohm’s law in honor of Georg Simon Ohm
(Fig. 4.1). The law clearly reveals that for a fixed resistance, the greater
the voltage (or pressure) across a resistor, the more the current, and the
more the resistance for the same voltage, the less the current. In other
words, the current is proportional to the applied voltage and inversely
proportional to the resistance.
By simple mathematical manipulations, the voltage and resistance
can be found in terms of the other two quantities:
E � IR
potential difference
���
resistance
V
I R
(volts, V) (4.3)
and R � � (ohms, �) (4.4)
E
I �
The three quantities of Eqs. (4.2) through (4.4) are defined by the
simple circuit of Fig. 4.2. The current I of Eq. (4.2) results from applying
a dc supply of E volts across a network having a resistance R. Equation
(4.3) determines the voltage E required to establish a current I
through a network with a total resistance R, and Equation (4.4) provides
the resistance of a network that results in a current I due to an
impressed voltage E.
Note in Fig. 4.2 that the voltage source “pressures” current in a
direction that passes from the negative to the positive terminal of the
battery. This will always be the case for single-source circuits. The effect
of more than one source in the network will be examined in the chapters
to follow. The symbol for the voltage of the battery (a source of
electrical energy) is the uppercase letter E, whereas the loss in potential
energy across the resistor is given the symbol V. The polarity of the
voltage drop across the resistor is as defined by the applied source
because the two terminals of the battery are connected directly across
the resistive element.
EXAMPLE 4.1 Determine the current resulting from the application
of a 9-V battery across a network with a resistance of 2.2 �.
Solution: Eq. (4.2):
E 9 V
I � � ���4.09 A
R 2.2 �
EXAMPLE 4.2 Calculate the resistance of a 60-W bulb if a current of
500 mA results from an applied voltage of 120 V.